IGCSE Physics Effects of Forces Questions and Worked Solutions (1.5.1) | 0625

By /
IGCSE Physics 0625 · Topic 1.5.1

IGCSE Physics Effects of Forces Questions and Worked Solutions

50 topic-wise practice questions on Effects of Forces (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions19 Core21 Extended10 Challenge121 marksTextbook §1.5 · pp. 33–54
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Effects of Forces

Forces change the shape, speed or direction of objects. When several forces act, their resultant decides what happens. This topic covers resultant forces, friction and air resistance, stretching springs, and (Extended) Newton's second law and circular motion.

Key formulas

F = m aresultant force F (N), mass m (kg), acceleration a (m/s²)Extended
k = F ÷ xspring constant k (N/m or N/cm), load F, extension xExtended
extension = stretched length − original lengthfor springsCore

Prerequisite knowledge

  • Speed and acceleration (1.2)
  • Mass and weight (1.3)
  • Plotting load–extension graphs

Exam tips & common mistakes

  • Always use the RESULTANT force in F = ma, not just the driving force
  • For a lift or rocket, resultant = upward force − weight
  • Convert cm to m if k is in N/m

Key facts from the syllabus (1.5.1)

  • A force can change an object's size, shape, speed or direction
  • Forces in the same direction add; opposite forces subtract to give the resultant
  • If the resultant force is zero, an object stays at rest or moves at constant velocity
  • Friction acts between surfaces and opposes motion, causing heating; air resistance (drag) is friction in a gas
  • Load–extension graph is a straight line through the origin up to the limit of proportionality
  • Extended the spring constant is force per unit extension
  • Extended an object moving in a circle at constant speed has a force towards the centre; more force is needed for higher speed, smaller radius or larger mass

Textbook: §1.5 Forces, pp. 33–54

1
Core[1 mark]

State two effects a force can have on an object.

Show solutionHide solution

Method: Forces change motion or shape.

Change its shape/size, or change its speed or direction
2
Core[2 marks]

Two forces act on a box along the same line: 30 N to the right and 18 N to the left. Calculate the resultant force.

Show solutionHide solution

Method: Subtract opposite forces.

R = 30 − 18 = 12 N to the right
3
Core[2 marks]

Forces of 12 N, 8 N (both to the right) and 15 N (to the left) act on an object. Calculate the resultant force.

Show solutionHide solution

Method: Add forces in the same direction, subtract opposite.

R = 12 + 8 − 15 = 5 N to the right
4
Core[2 marks]

A car moves at constant speed in a straight line. State the resultant force on it and explain.

Show solutionHide solution

Method: No resultant force means no change in velocity.

Zero; the driving force balances the resistive forces
5
Core[2 marks]

State two ways a resultant force can change the velocity of an object.

Show solutionHide solution

Method: Velocity has speed and direction.

It can change the speed or change the direction of motion
6
Core[2 marks]

Describe what is meant by friction between two solid surfaces.

Show solutionHide solution

Method: Friction opposes relative motion and produces heating.

A force between two surfaces that impedes motion and causes heating
7
Core[1 mark]

Name the frictional force on an object moving through air.

Show solutionHide solution

Method: Drag in a gas.

Air resistance (drag)
8
Core[3 marks]

Describe an experiment to plot a load–extension graph for a spring.

Show solutionHide solution

Method: Hang spring from clamp, measure original length with ruler, add masses one at a time, measure new length, extension = new − original.

Add known loads, measure the length each time with a metre rule, calculate extension = length − original length, plot load against extension
9
Core[2 marks]

A spring is 12.0 cm long with no load and 15.6 cm long with a 2.0 N load. Calculate the extension.

Show solutionHide solution

Method: Extension = loaded length − original length.

x = 15.6 − 12.0 = 3.60 cm
10
Core[2 marks]

State what is shown by a straight line through the origin on a load–extension graph.

Show solutionHide solution

Method: Direct proportion.

Extension is directly proportional to the load
11
Core[2 marks]

Forces of 50 N to the right and 50 N to the left act on a box. State the resultant force and describe the motion if it was moving.

Show solutionHide solution

Method: Balanced forces.

Zero; it continues at constant velocity
12
Core[2 marks]

A spring has an unstretched length of 8.0 cm. With a 4.0 N load its length is 11.2 cm. Calculate the extension.

Show solutionHide solution

Method: Length − original.

x = 3.20 cm
13
Core[2 marks]

State what happens to a spring if it is stretched beyond its limit of proportionality.

Show solutionHide solution

Method: No longer proportional.

Extension is no longer proportional to load (the graph curves) and it may not return to its original length
14
Core[3 marks]

A load–extension table gives 1.0 N: 2.0 cm, 2.0 N: 4.0 cm, 3.0 N: 6.0 cm, 4.0 N: 8.5 cm. State whether Hooke's law is obeyed throughout.

Show solutionHide solution

Method: Check proportionality.

Obeyed up to 3.0 N; at 4.0 N the extension is more than 8.0 cm, so the limit of proportionality has been passed
15
Core[2 marks]

Explain why a cyclist must keep pedalling to maintain a constant speed.

Show solutionHide solution

Method: Friction and air resistance.

Pedalling provides a driving force that balances friction and air resistance; without it there is a resultant backward force
16
Core[2 marks]

State two ways friction can be useful and one way it is a nuisance.

Show solutionHide solution

Method: Examples.

Useful: brakes, walking/tyre grip. Nuisance: wear and heating in engines
17
Core[2 marks]

A car's driving force is 2000 N and resistive forces are 1500 N. Calculate the resultant force.

Show solutionHide solution

Method: Subtract.

R = 500 N forwards
18
Core[1 mark]

State the effect of a resultant force on a stationary object.

Show solutionHide solution

Method: Accelerates.

It starts to move (accelerates) in the direction of the force
19
Core[2 marks]

Explain why air resistance on a car increases as its speed increases.

Show solutionHide solution

Method: Collisions with air.

At higher speed the car hits more air particles each second with greater force
20
Extended[1 mark]

Define the spring constant.

Show solutionHide solution

Method: k = F/x.

Force per unit extension
21
Extended[2 marks]

A spring extends by 4.0 cm when a 6.0 N force is applied. Calculate the spring constant in N/cm.

Show solutionHide solution

Method: k = F ÷ x.

k = 6.0 ÷ 4.0 = 1.50 N/cm
22
Extended[2 marks]

A spring has k = 40 N/m. Calculate the extension produced by a 3.0 N load.

Show solutionHide solution

Method: x = F ÷ k.

x = 3.0 ÷ 40 = 0.0750 m = 7.50 cm
23
Extended[1 mark]

State what is meant by the limit of proportionality.

Show solutionHide solution

Method: Beyond this point, the graph is no longer a straight line.

The point beyond which extension is no longer proportional to load
24
Extended[2 marks]

A resultant force of 2400 N acts on a car of mass 1200 kg. Calculate the acceleration.

Show solutionHide solution

Method: F = ma so a = F/m.

a = 2400 ÷ 1200 = 2.00 m/s²
25
Extended[2 marks]

Calculate the resultant force needed to accelerate a 0.15 kg ball at 40 m/s².

Show solutionHide solution

Method: F = ma.

F = 0.15 × 40 = 6.00 N
26
Extended[3 marks]

A rocket of mass 500 kg has a thrust of 7400 N upwards. Calculate its acceleration (g = 9.8 N/kg).

Show solutionHide solution

Method: Resultant = thrust − weight.

W = 4900 N; R = 7400 − 4900 = 2500 N; a = 2500 ÷ 500 = 5.00 m/s² upwards
27
Extended[1 mark]

State the direction of the acceleration produced by a resultant force.

Show solutionHide solution

Method: F = ma: same direction.

The same direction as the resultant force
28
Extended[3 marks]

An object moves in a circle at constant speed. Explain why there must be a resultant force on it.

Show solutionHide solution

Method: Its direction (and so velocity) keeps changing.

Its velocity changes direction, so it accelerates; this needs a resultant force towards the centre (perpendicular to motion)
29
Extended[2 marks]

A ball on a string moves in a circle. The tension is increased while mass and radius stay the same. State what happens to the speed.

Show solutionHide solution

Method: More centripetal force at the same radius allows a higher speed.

The speed increases
30
Extended[2 marks]

A spring has spring constant 25 N/cm. Calculate the load that produces an extension of 3.0 cm.

Show solutionHide solution

Method: F = kx.

F = 25 × 3.0 = 75.0 N
31
Extended[2 marks]

State what the gradient of a load–extension graph represents (load on y-axis).

Show solutionHide solution

Method: k = F/x.

The spring constant
32
Extended[3 marks]

Two identical springs each have k = 20 N/m. One spring is stretched by 0.15 m. Calculate the force needed and the extension of the same spring with a 1.5 N load.

Show solutionHide solution

Method: F = kx.

F = 20 × 0.15 = 3.00 N; x = 1.5 ÷ 20 = 0.0750 m
33
Extended[2 marks]

A 0.50 kg trolley accelerates at 1.6 m/s². Calculate the resultant force.

Show solutionHide solution

Method: F = ma.

F = 0.50 × 1.6 = 0.800 N
34
Extended[3 marks]

A 1500 kg car's driving force is 4500 N and the resistive force is 1200 N. Calculate its acceleration.

Show solutionHide solution

Method: R = 3300 N.

a = 3300 ÷ 1500 = 2.20 m/s²
35
Extended[3 marks]

A lift of mass 800 kg accelerates upwards at 0.50 m/s². Calculate the tension in the cable (g = 9.8 N/kg).

Show solutionHide solution

Method: T − mg = ma.

T = 800 × 9.8 + 800 × 0.50 = 8240 N
36
Extended[2 marks]

A resultant force of 12 N gives an object an acceleration of 3.0 m/s². Calculate its mass.

Show solutionHide solution

Method: m = F ÷ a.

m = 12 ÷ 3.0 = 4.00 kg
37
Extended[3 marks]

A car moving in a circle at constant speed: state the direction of the resultant force and explain what would happen if it suddenly became zero.

Show solutionHide solution

Method: Centripetal.

Towards the centre of the circle; the car would move off in a straight line (along a tangent)
38
Extended[2 marks]

For an object moving in a circle, state what happens to the force needed if the speed increases at the same radius and mass.

Show solutionHide solution

Method: F increases.

A larger force is needed
39
Extended[2 marks]

For an object moving in a circle at constant speed, state what happens to the force needed if the radius decreases.

Show solutionHide solution

Method: F increases.

A larger force is needed
40
Extended[2 marks]

For an object moving in a circle at constant speed and radius, state what happens to the force needed if the mass increases.

Show solutionHide solution

Method: F increases.

A larger force is needed

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A spring obeys Hooke's law up to 6.0 N. Its natural length is 10.0 cm and it is 13.0 cm long with a 2.0 N load. Calculate its length with (a) a 5.0 N load, (b) a 0.50 N load.

Show solutionHide solution

Method: Extension per N = 1.5 cm.

(a) 10.0 + 7.5 = 17.5 cm (b) 10.0 + 0.75 = 10.75 cm
42
ChallengeCore[4 marks]

A car travels at constant speed while the engine provides 3200 N. (a) State the total resistive force. (b) The driving force increases to 4000 N. Describe and explain the motion.

Show solutionHide solution

Method: Balanced, then unbalanced.

(a) 3200 N (b) There is a resultant forward force of 800 N, so it accelerates; as speed increases, air resistance increases until forces balance again at a higher speed
43
ChallengeCore[3 marks]

Describe how friction is both a problem and essential for a car moving on a road.

Show solutionHide solution

Method: Examples.

Essential: tyre grip and brakes; problem: friction in moving parts wastes energy as heat and causes wear
44
ChallengeExtended[4 marks]

Two springs, each with k = 40 N/m, are joined end to end (in series). A 2.0 N load hangs from the bottom. Calculate the total extension and the effective spring constant.

Show solutionHide solution

Method: Each spring carries 2.0 N.

Each extends 0.050 m; total = 0.100 m; k = 2.0 ÷ 0.10 = 20.0 N/m
45
ChallengeExtended[4 marks]

The same two springs (k = 40 N/m each) now hang side by side sharing a 2.0 N load equally. Calculate the extension and the effective spring constant.

Show solutionHide solution

Method: Each carries 1.0 N.

x = 1.0 ÷ 40 = 0.0250 m; k = 80.0 N/m
46
ChallengeExtended[5 marks]

A 1200 kg car accelerates from rest to 20 m/s in 10 s while a constant resistive force of 600 N acts. Calculate (a) the acceleration, (b) the resultant force, (c) the driving force.

Show solutionHide solution

Method: a = Δv/Δt; F = ma.

(a) 2.00 m/s² (b) 2400 N (c) 2400 + 600 = 3000 N
47
ChallengeExtended[4 marks]

A 70 kg skydiver has an acceleration of 4.0 m/s² downwards at one moment (g = 9.8 N/kg). Calculate the air resistance at that moment.

Show solutionHide solution

Method: mg − F = ma.

F = 70 × (9.8 − 4.0) = 406 N
48
ChallengeExtended[4 marks]

A rocket of mass 2.0 × 10⁴ kg has an upward thrust of 3.0 × 10⁵ N (g = 9.8 N/kg). Calculate its initial acceleration and explain why the acceleration increases as it rises.

Show solutionHide solution

Method: Resultant = thrust − weight.

a = (3.0×10⁵ − 1.96×10⁵) ÷ 2.0×10⁴ = 5.20 m/s²; it burns fuel, so its mass (and weight) decreases
49
ChallengeExtended[4 marks]

A car goes round a roundabout at constant speed. Explain why it is accelerating, state the direction of the resultant force, and name the force that provides it.

Show solutionHide solution

Method: Changing direction.

Its velocity changes direction, so it accelerates; the resultant force acts towards the centre; it is provided by friction between the tyres and the road
50
ChallengeExtended[3 marks]

A stone on a string is whirled in a horizontal circle. Explain what happens to the tension needed if the stone is whirled faster in the same circle, and what happens if the string breaks.

Show solutionHide solution

Method: More force needed.

More tension is needed at higher speed; if the string breaks, the stone moves off in a straight line along the tangent
All IGCSE Physics topics← 1.4 Density1.5.2 Turning Effect of Forces →

Book a free trial class

Tell us a little about your requirement.